Finding Subspaces that Satisfy Specific Intersections in \mathbb{R}^3

[moonbeam]

Homework Statement

Find subspaces A, B, and C of \mathbb{R}^3 so that A \cap B \cap C \ne \{\vec{0}\} and (A + B) \cap C \ne A \cap C + B \cap C.
You can specify a subspace by the form A = span\{\vec{e}_1, \vec{e}_2\}.

Homework Equations

A + B is the set of all vectors in \mathbb{R}^3 of the form \vec{a} + \vec{b}, where \vec{a} is in A and \vec{b} is in B.

The Attempt at a Solution

I’ve been trying to guess. I’ve tried A as a line, B as a plane containing A, and C as \mathbb{R}^3 itself. I’ve tried making all three subspaces planes that intersect in a line. I’ve tried making A and B planes that intersect in a line and C as \mathbb{R}^3 itself. I’ve tried making A equal to B and C another plane intersecting them in a line.
Finding A \cap B \cap C \ne \{\vec{0}\} is not my problem. I’m having trouble satisfying the second specification: (A + B) \cap C \ne A \cap C + B \cap C.

 

[ZioX]

if A and B are subspaces then so is A+B. Can you find a basis of A+B knowing a basis of A and B? How does this help?

We also know that the intersection of subspaces is a subspace.

 

[moonbeam]

I can't find the bases without knowing how many dimensions each subspace is. How do I even know what dimension to make the subspaces A, B, and C?

 

[uiulic]

Check my awkward method below,

Imagine stuff like Cartesian coordinate systems, then we have X, Y,Z axis(orthongol to each other).The R^3 can be expressed by points in OXYZ, with XZ plane at the bottom.

XY plane and bottom make a line x. YZ plane and bottom make a line z.[WHY I CANNOT COPY YOUR EQUATIONS AND JUST PASTE THEM HERE?]

According to definition of XY+YZ, the set has another POINT OTHER THAN x line and z line , but that point, e.g. point (1,0,1)=(1,0,0)+(0,0,1) ,lie on the bottom plane.


Or using set language

A={(x,y,z);x=0}, B={(x,y,z);z=0}, C={(x,y,x);y=0}

A and C make {(x,y,z);x=0,y=0}, B and C make {(x,y,z);z=0,y=0};

(A+B) and C includes a point {(1,0,1)},This compltes your problem.

 

[moonbeam]

A={(x,y,z);x=0}, B={(x,y,z);z=0}, C={(x,y,x);y=0}

A and C make {(x,y,z);x=0,y=0}, B and C make {(x,y,z);z=0,y=0};

(A+B) and C includes a point {(1,0,1)},This compltes your problem.

In this situation the subspaces A, B, and C do not satisfy the first requirement: A \cap B \cap C \neq \{\vec{0}\}. The only vector in common among the 3 planes is the zero vector. But the problem is asking for subspaces A, B, and C, that intersect in more than just the case of the zero vector.

 

[uiulic]

moonbeam,

I had thought your questions are separate (since you said meeting the first one is not your problem).

But it is handy as well. Shift the bottom to one unit.See below

A={(x,y,z);x=0}, B={(x,y,z);z=0}, C={(x,y,z);y=1}

A and C make {(x,y,z);x=0,y=1}, B and C make {(x,y,z);z=0,y=1};

(A+B) and C includes a point (1,1,1),This compltes your problem

 

[moonbeam]

moonbeam,

I had thought your questions are separate (since you said meeting the first one is not your problem).

But it is handy as well. Shift the bottom to one unit.See below

A={(x,y,z);x=0}, B={(x,y,z);z=0}, C={(x,y,z);y=1}

A and C make {(x,y,z);x=0,y=1}, B and C make {(x,y,z);z=0,y=1};

(A+B) and C includes a point {(1,1,1)},This compltes your problem

The point (1, 1, 1) is not the same as the vector <1, 1, 1>. If the planes only meet at one particular point, then their intersection is just the zero vector. So the intersection of these 3 planes is still just the zero vector. I need the 3 planes to intersect in a line segment, not just one point.

 

[uiulic]

The point (1, 1, 1) is not the same as the vector <1, 1, 1>. If the planes only meet at one particular point, then their intersection is just the zero vector. So the intersection of these 3 planes is still just the zero vector. I need the 3 planes to intersect in a line segment, not just one point.

Where is your statement from? I haven't touched mathematics for many years, but I believe your understanding is completely wrong.

A={(x,y,z);x=0}, B={(x,y,z);z=0}, C={(x,y,z);y=1}

A AND B AND C={(0,1,0)} , IS THIS A ZERO VECTOR IN R^3? You may be confused by the point(working as a vector) with the vector with magnitude and direction stuff? That can be called a located vector (by serge lang), which can be taken as an ordered pair of points.In vector space (subspace), points (if you think of it as an element in your R^3) ARE vectors. Also, don't get confused by the vector used in tensor analysis. Vector space is a set. It is easier to use some set language to help your analysis. But vector space can be an element in another set. What really is an element, or a set, or a vector space in your analysis of the problem, is your decision depending on what kind of problems ;and once you decide (e.g. a point is a vector), then the analysis must be consistent. Since all your statements are sets in R^3, then A,B,C ARE SETS including many elements (each element belongs to R^3, and each element can be expressed by an ordered triple or a point with the three coordinates). That's why I write them as written.Noting I embedded a coordinate system to help our analysis. Under such a coordinate system, the problem can be easily described. Now, I only need to find one case that DISobey the corresponding assertion.

A and C make {(x,y,z);x=0,y=1}, B and C make {(x,y,z);z=0,y=1};

(make sure you know what A+B is about according to its definition, also make sure THE ELEMENTS IN A+B IS STILL POINTS OR an ordered triple WITH THREE COORDINATES )
(A+B) and C includes a point (1,1,1), (this is just an alternative to equation problem, but this can be obtained without resorting to equations, only need to make a big guess as your case is simple). This point will be neither in {(x,y,z);x=0,y=1} nor in {(x,y,z);z=0,y=1} (this is obvious). This compltes your problem

 

[moonbeam]

This is what I eventually came up with and I got confirmation from a member of another math forum.

Let A = span\{\left[\begin{array}{c}1\\0\\0\end{array}\right] \left[\begin{array}{c}0\\0\\1\end{array}\right]\}.
Let B = span\{\left[\begin{array}{c}0\\1\\0\end{array}\right] \left[\begin{array}{c}0\\0\\1\end{array}\right]\}.
Let C = span\{\left[\begin{array}{c}1\\1\\0\end{array}\right] \left[\begin{array}{c}0\\0\\1\end{array}\right]\}.

Then A \cap B \cap C = span\{\left[\begin{array}{c}0\\0\\1\end{array}\right]\},
(A+B) \cap C = span\{\left[\begin{array}{c}1\\1\\0\end{array}\right] \left[\begin{array}{c}0\\0\\1\end{array}\right]\},
and A \cap C + B \cap C = span\{\left[\begin{array}{c}0\\0\\1\end{array}\right]\}.

\therefore (A + B) \cap C \ne A \cap C + B \cap C.